log problem: solve 2lnx + 5ln2 = lnx - ln4

[Guest]

Solve the following equation for the unknown variable. Justify each step of the solution with a written explanation. Round to the nearest thousandth:
2lnx+5ln2=lnx-ln4

 

[stapel]

You need to use log rules:

. . . . .log_b(aⁿ) = nlog_b(a)

. . . . .log_b(ac) = log_b(a) + log_b(c)

. . . . .log_b(a/c) = log_b(a) - log_b(c)

First, put the multipliers on the left-hand side inside their respective log terms. Then combine the log terms on the left- and right-hand sides, to get "ln(something) = ln(something else)".

Then you can equate the "something" with the "something else", and solve the resulting quadratic equation.

If you get stuck, please reply showing how far you have gotten. Thank you.

Eliz.

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
2\ln (x) + 5\ln (2) & = & \ln (x) - \ln (4) \\
\ln (x) & = & - \ln (32) - \ln (4) \\
\ln (x) & = & - \ln (8) \\
\end{array}\)

 

[stapel]

I could be wrong, but:

. . . . .-ln(32) - ln(4) = -[ln(32) + ln(4)] = -[ln(128)]

...not -ln(8).

Eliz.

 

[pka]

I could be wrong, but:

. . . . .-ln(32) - ln(4) = -[ln(32) + ln(4)] = -[ln(128)]
...not -ln(8).Eliz.

No you are correct! Sometimes I just jump!
This time I forgot the minus sign.
Thanks for checking!